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3 years ago

Pleaseeee help meee i dont understand

Mathematics

1 answer:

prohojiy [21]3 years ago

3 0

-(5y) - y = 18
-6y = 18
y = -3
x = 5(-3)
x = -15
(-15, -3)

5x + 4(-7x - 3) = -35
5x -28x -12 = -35
-23x = -23
x = 1
y = -7(1) - 3
y = -10
(1,-10)

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The number of pages read by two students each day for 5 days are listed in the table. Which statements are true about the data s

monitta

Answer:

The means for both data sets are equal.
The range of the number of pages Lionel read is greater than the range of the number of pages Aaron read.
The range of the number of pages that Lionel read is 75.

Step-by-step explanation:

Given

Lionel

90, 45, 30, 25, 15

Aaron

55, 19, 40, 16, 75

in the first case; For Lionel;

Mean = (90+45+30+25+15 )/5 = 41 and

Range = (90-15)=75

The second case; Aaron

Mean = (55+19+40+16+75)/5 = 41 and

Range = (75-16)= 59

Therefore;

Both data sets have the same mean but different range.

The range of the number of pages Lionel is 75, which is greater than the range of the number of pages Aaron read, which is 59.

6 0

3 years ago

Read 2 more answers

Deva bought 9 1/2 m of cloth for rupees 28 1/2.Samruth buys 26 1/3 m of same cloth. how much should samruth pay the shopkeeper?

likoan [24]

For this case we can make the following rule of three:

9 1/2 m --------------> 28 1/2 rupees

26 1/3 m -------------> x

From here, we clear the value of x.

We have then:

$x = (\frac{26+\frac{1}{3}}{9+\frac{1}{2}}) (28+\frac{1}{2})$

Rewriting we have:

$x = (\frac{\frac{79}{3}}{\frac{19}{2}}) (\frac{57}{2})$

$x = (\frac{158}{57}) (\frac{57}{2})$

$x = \frac{158}{2}$

$x = 79$

Answer:

samruth should pay 79 rupees to the shopkeeper

5 0

3 years ago

A couple book a cruise to Alaska that promises to refund 100 per day of rain on the seven day cruise up to a maximum of 300. The

zubka84 [21]

Answer:

the variance of the refund payment to the couple = 9463.394

Step-by-step explanation:

Given that :

A couple book a cruise to Alaska that promises to refund 100 per day of rain on the seven day cruise up to a maximum of 300.

It is possible that the couple won't be able to refund up 100 per day or more than 100 per day.

SO; let assume that the refund payment happens to be 0, 100,200, 300

Let X be the total refund payment on the seven day cruise.

We can say X = 0, if there is no rain on all 7 days.

$P(X = 0) = _nC_x * P^x * (1 - P)n-x$

$P(X = 0) = _7C_o * 0.2^0 * (1-0.2)^{7-0$

$P(X = 0) =1 * 1* (1-0.2)^{7$

$P(X = 0) =(0.8)^{7$

$P(X = 0) =0.2097152$

If it rains on any one day; then X = 100

$P(X = 100) = _nC_x * P^x * (1 - P)n-x$

$P(X = 100) = _7C_1 * 0.2^1 * (1-0.2)^{7-1$

$P(X = 0) =7 * 0.2* (1-0.2)^{6$

$P(X = 100) =7* 0.2* (0.8)^{6$

$P(X = 100) =0.3670016$

if it rains on any two day ; then X = 200

$P(X = 200) = _nC_x * P^x * (1 - P)n-x$

$P(X = 200) = _7C_2 * 0.2^2 * (1-0.2)^{7-2$

$P(X = 200) = 21 * 0.2^2 * (0.8)^{5$

$P(X = 200) = 0.2752512$

if it rains on any three day or more than that ; then X = 300

$P(X \ge 300) = 1 - P(X < 300) \\ \\ P(X \ge 300) = 1 - [P(X = 0) + P(X = 100) + P(X = 200)] \\ \\ P(X \ge 300) = 1 - [0.2097152 + 0.3670016 + 0.2752512] \\ \\ P(X \ge 300) = 0.148032$

Now; we have our probability distribution function as:

P(X = 0) = 0.2097152

P(X = 100) = 0.3670016

P(X = 200) = 0.2752512

P(X = 300) = 0.148032

In order to determine the variance of the refund payment to the couple; we use the formula:

variance of the refund payment to the couple $[Var X] =E [X^2] - (E [X])^2$

where;

$E[X^2] = \sum x^2 \times p \\ \\ E[X^2] = 0^2 * 0.2097152 + 100^2 * 0.3670016 + 200^2 * 0.2752512 + 300^2 * 0.148032 \\ \\ E[X^2] = 0 + 3670.016 + 11010.048+ 13322.88 \\ \\ E[X^2] =28002.944$

$(E [X]) = \sum x * p\\ \\ (E [X]) = 0 * 0.2097152 + 100 * 0.3670016 + 200 * 0.2752512 + 300 * 0.148032 \\ \\ (E [X]) = 0 + 36.70016 + 55.05024 + 44.4096\\ \\ (E [X]) = 136.16 \\ \\ (E [X])^2 = 136.16^2 \\ \\ (E [X])^2 = 18539.55$